Minimum of an Octic

$f(t) = (t^2 - 3)(t^2 - 9)(t^2 - 5)(t^2 - 7) + 100 = (t^4 - 12t^2 + 27) (t^4 - 12t^2 + 35) + 100$ .

Let $k = t^4 - 12t^2 + 36$ , then $k = (t^2 - 6)^2 \geq 0$ . Thus

$f(t) = (k - 9)(k - 1) + 100 = k^2 -10k + 109$ $= (k^2 - 10k + 25) + 84 = (k - 5)^2 + 84 \geq 84$

Hence, the minimum is 84, which is achieved when $k = 5$ or $t^2 = \pm \sqrt(5) + 6$ .

If you make another variable x, and set $x = t^2 - 6$ , by substituting in x you get $(x+3)(x+1)(x-1)(x-3) + 100$ $(x^2 - 9)(x^2 +9) + 100$ $x^4 - 10x^2 + 109$ By treating this as a quadratic you can get that the minimum for this expression is when $x^2 = \frac {10}{2}$ $25 - 50 + 109 = 84$

$f(t) - 100 = (t^4 - 12 t^2 + 27) (t^4 - 12 t^2 + 35) = (t^4 - 12t^2 + 31)^2 - 4^2$ , so $f(t) \geq 100 - 16 = 84$ . We have equality when $t^4 - 12t^2 + 31 = 0$ , which occurs when $t^2 = 6 \pm \sqrt{5}$ .

Note: The minimum doesn't occur when $t^2=6$ , which might be a common guess and which gives an answer of 109.

(t²-3)(t²-5)(t²-7)(t²-5) = f(t)

if t²-5=k

k(k+2)(k-2)(k-4) (k²-2k)²-8(k²-2k) fatored

if k²-2k=z

z²-8z dt/d 2z-8=0

z=4

4²-8*4=-16

100+(-16)=84

C.Q.D

The expression in blue is of the form $x^2 \ge 0$ .

$\begin{aligned} f(t) &= 100 + (t^2-3)(t^2-9)\cdot(t^2-5)(t^2-7) \\ &= 100 + (t^2-6+3)(t^2-6-3)\cdot(t^2-6+1)(t^2-6-1) \\ &= 100 + ((t^2-6)^2 -9)((t^2-6)^2-1) \\ &= 100 + ((t^2-6)^2 -5-4)((t^2-6)^2-5+4) \\& = 100 + {\color{#3D99F6}{((t^2-6)^2-5)^2}} - 16 \\ &\ge 100 -16 \\ &= 84 \end{aligned}$

It is also clear that the minimum is reached when $(t^2-6) - 5 = 0$ , which implies $t = \pm\sqrt{6 \pm\sqrt5}$ .

Let $k = (t^2-6)$

$f(k) = (k + 3)(k + 1)(k - 1)(k - 3) + 100$

$f(k) = (k^2 - 9)(k^2 - 1) + 100$

Locate the minimum using calculus $f'(k) = 4k^3 - 20k = 0 => 4k^2 = 20 => k=\sqrt5$

$f(\sqrt5) = (5 - 9)(5-1) + 100 = (-4)(4) + 100 = -16 + 100 = 84$

In order to have the multiplication of the four brackets to be a -tive integer~~, $value~ would~ be ~either~~t^2= 4 ~~or~~t^2= 8. ~\\ We~ see~ that~~~3<t^2<4 ~~or~~ >8<t^2<9~~ gives~~ us ~~product~of ~the~brackets~>15,~~a~value~ obtained~by~t^2=4~or~8.\\ Investigating~ values~ near~ and ~less~than~4, and~values~ near~ and ~greater~than~8,~we ~get~f(t)= ~-~16.\\ So~f(t)_{min}=100-16=84.$

$f(t) = (t^{2} - 3)(t^{2} - 5)(t^{2} - 7)(t^{2} - 9) + 100 = (t^{4} - 12t^{2} + 27)(t^{4} - 12t^{2} + 35) + 100$ Let us assume $t^{4} - 12t^{2} + 27 = k$

Thus f(t) becomes : $k(k+8) + 100$ Now the minimum value of this quadratic is given by -(64-400)/4 = 84 Hence the answer is 84.

A quick inspection of the problem will show the solution is less that 100 for 3 < t^2 < 5. Finding a minimum will require differentiation but without the right substitute, setting the differential to zero and then solving the resultant polynomial of order 7 will be extremely difficult. Let p = (t^2 - 3)(t^2 - 9) and let q = (t^2 - 5)(t^2 - 7) so f(t) = pq p= t^4 - 12t^2 + 27 and q = t^4 - 12t^2 + 35 which gives q = p + 8. f(t) = p(p+8) = p^2 + 8p At this point we can differentiate using the chain rule with df/dt = df/dp * dp/dt

df/dp = 2p + 8 = 2(p + 4) dp/dt = 4t^3 - 24t = 4t(t^2 - 6)

df/dt = 2(p + 4)*4t(t^2 - 6) = 8t(t^4 -12t^2 + 31)(t^2 - 6)

At a minimum, df/dt = 0 giving us 3 sets of solutions: 8t = 0, t^4 -12t^2 + 31 = 0 and t^2 - 6 = 0. From our initial inspection of the problem and without much calculation we can see that the minimum must be a solution within the middle of these 3 sets of solutions as 8t = 0 and t^2 - 6 = 0 will both give values of f(t) > 100. As t^4 -12t^2 + 31 is a quadratic in t^2 we can use the quadratic formula. Hence, t^2 = (12 +-sqrt20) / 2 = 6 +-sqrt5

Substituting either the positive or the negative root version of t^2 here will result in the same value of f. Here i've used the negative root: f(t) = (t^2 - 3)(t^2 - 5)(t^2 - 7)(t^2 - 9) + 100 = (3 - sqrt5)(1 - sqrt5)(-1 - sqrt5)(-3 - sqrt5) + 100 = (8 - 4sqrt5)(8 + 4sqrt5) +100 = 64 - 80 +100 =84

f(t) = (t^2−3)(t^2−5)(t^2−7)(t^2−9)+100 f(t) = t^8 -24t^6 + 206t^4 - 744t^2 + 1045

let derived to be ...

f'(t) = 8t^7 - 144t^5 + 824t^3 - 1488t = 0

First we will factor our t: 8t^6 - 144t^4 + 824t^2 - 1488 = t ( 8t^6 - 144t^4 + 824t^2 - 1488 ). t = 0 is a root of multiplicity 1. Now we will find remaining roots using polynomial 8t^6 - 144t^4 + 824t^2 - 1488.

t1 = 0 (max) t2 = 1.94009 (min) t3 = -1.94009 (min) t4 = -2.44949 (max) t5 = 2.44949 (max) t6 = -2.86986 (min) t7 = 2.86986 (min)

then t2 or t3 or t6 or t7 inserted into the equation early. And the result is 84